2x Examine whether the value of variable inserted within the brackets is the root of the 5. given equation or not. + 5 = 7, (x = 3) (iii) x - 4 = 1, (x = 3) (i) 4x + 3 = 7, (x = 1) (11) (v) 5x – 1 = 7, (x = 2) (vi) x + 9 = 13, (x = 4) (iv) 6x = 18, (x = 2) , у (vii) 3 + 5 = 8, (y = 9) (ix) + 4 = 5, (p = 1) (vii) 5x – 7 = 8, (x = 3) 5 , ) = , () (x) * = 6, (x = 42) : 6, = 3 = - = р 5 X = 7 .
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Verified answer
Step-by-step explanation:
i) x + 3 = 0
By putting x = 3
L.H.S. = 3 + 3 = 6 ≠ R.H.S.
Hence, the equation is not satisfied.
(ii) x + 3 = 0
By putting x = 0
L.H.S. = 0 + 3 = 3 ≠ R.H.S.
Hence, the equation is not statisfied.
(iii) x + 3 = 0
By putting x = – 3
L.H.S. = – 3 + 3 = 0 = R.H.S.
Hence, the equation is satisfied.
(iv) x – 7 = 1
By putting x = 7
L.H.S. = 7 – 7 = 0 ≠ R.H.S.
Hence, the equation is not satisfied.
(v) x – 7 = 1
By putting x = 8
L.H.S. = 8 – 7 = 1 = R.H.S.
Hence, the equation is satisfied.
(vi) 5x = 25
By putting x = 0
L.H.S. = 5 × 0 = 0 ≠ R.H.S.
Hence, the equation is not satisfied. (vii) 5x = 25
By putting x = 5
L.H.S. = 5 × 5 = 25 = R.H.S.
Hence, the equation is satisfied.
(viii) 5x = 25
By putting x = – 5
L.H.S. = 5 × (–5) = –25 ≠ R.H.S.
Hence, the equation is not satisfied. (ix) m/3=2
By putting m = – 6
L.H.S.R.H.S.=−=−≠632
Hence, the equation is not satisfied.
(x) m/3=2
By putting m = 0
L.H.S. = 03 = 0 ≠ R.H.S.
Hence, the equation is not satisfied.
(xi) m/3= 2
By putting m = 6
L.H.S. = 63 = 2 = R.H.S
Hence, the equation is satisfied.
Q 2.
Check whether the value given in the brackets is a solution to the given equation or not:
(A) n + 5 = 19 (n = 1)
(B) 7n + 5 = 19 (n = –2)
(C) 7n + 5 = 19 (n = 2)
(D) 4p – 3 = 13 (p = 1)
(E) 4p – 3 = 13 (p = – 4)
(F) 4p – 3 = 13 (p = 0)